Hibernate Subselect with class inheritance

Question

I have two objects with same fields, but from different tables (foo_something and bar_something in fact not plain tables, there are results of several join operations). I use @Subselect hibernate annotation for fetching data. Can I make class structure like this, but which working fine? This class structure doesn't work because annotations doesn't inherit. If I use @Inheritance annotation, I get error Unknown column foo_somethingca0.DTYPE

class Something {

    @Id @Column Long id;
    @Column String name;
    @Column String description;

    //getters, setters, constructors

}

@Entity
@Immutable
@Subselect("SELECT id, name, description FROM foo_something")
class FooSomething extends Something {

}

@Entity
@Immutable
@Subselect("SELECT id, name, description FROM bar_something")
class BarSomething extends Something {

}

UPD.

Thanks, @HenryMartens, I added annotations @Entity and @Inheritance with TABLE_PER_CLASS strategy, and made class abstract, now it's working fine:

@Entity
@Inheritance(strategy = TABLE_PER_CLASS)
abstract class Something {

    @Id @Column Long id;
    @Column String name;
    @Column String description;

    //getters, setters, constructors

}

@Entity
@Immutable
@Subselect("SELECT id, name, description FROM foo_something")
class FooSomething extends Something {

}

@Entity
@Immutable
@Subselect("SELECT id, name, description FROM bar_something")
class BarSomething extends Something {

}

score 1 · Accepted Answer · 2017-08-28 12:29:47Z

1

The DTYPE in Hibernate is the discriminator column. It is used to tell the difference between some objects that are persisted in the same database table like your example. You can add an discriminator column with the @DiscriminatorColumn annotation.

On the concrete classes you can use @DiscriminatorValue to set the value of the discriminator column for the annotated class.

For example:

@Entity
@Inheritance( strategy = InheritanceType.TABLE_PER_CLASS ) 
@DiscriminatorColumn(name="DISC", discriminatorType=STRING, length=20)
class abstract Something {

    @Id 
    @Column 
    Long id;

    @Column 
    String name;

    @Column 
    String description;

    //getters, setters, constructors 
}

@Entity
@Table("FOO_SOMETHING")
@Immutable
@DiscriminatorValue("FooSomething")
class FooSomething extends Something {

}

@Entity
@Table("BAR_SOMETHING")
@Immutable
@DiscriminatorValue("BarSomething")
class BarSomething extends Something {

}

edited Aug 28, 2017 at 12:29

answered Aug 28, 2017 at 11:58

user8227693

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2 Comments

dgabriel Over a year ago

My base class Something - it's not a table, it's simple base class with list of fields, getters and setters, which i don't want to duplicate in BarSomething and FooSomething, that's why I cannot use @Table annotation in base class cause SchemaManagementException: Schema-validation: missing table [SOMETHING]. This class an abstract.

user8227693 Over a year ago

@Denis Gavrus Then you have to make your Something class abstract. You can use InheritanceType.TABLE_PER_CLASS to get a table for each FooSomething and BarSomething.

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